L'Hospital's Rule: A Proof of the Infinity over Infinity Case
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- Опубліковано 8 лют 2025
- This is an intuitive proof of l'Hospital's rule in the infinity over infinity case.
For a fully rigorous proof, click on the following link: drive.google.c...
Course Web Page: sites.google.c...
How does f’(c)/g’(c) approach L if only x approaches a and y is fixed? Couldn’t c remain close to y and therefore not approach a?
I have added a link in the description of this video to a document that contains a complete and fully rigorous proof of the result. Cheers!
@@slcmathpc Hello, I'm not a mathematician (nor a maths student) so maybe what I'm saying is nonsense but I'm reading through the rigorous proof and I still don't understand how we can just fix y to be a constant. If y is in the interval (a - delta 1; a + delta 1) and it is also fixed, then if we choose an epsilon that's small enough, the interval will get too narrow to include y (because y stays in place and the interval shrinks with decreasing epsilon). So I don't understand how y can be fixed and also always be in the interval (a - delta 1; a + delta 1).
Your confusion stems from the fact that you are not yet fully comfortable with the rigorous (epsilon/delta) notion of convergence. To show that f(x)/g(x) converges to L, as x approaches a, we have to show that for any small epsilon>0, we can find a small interval around a, say (a-delta,a+delta), where delta>0, such that |f(x)/g(x)-L| < epsilon, if x is in (a-delta,a+delta). In our case, the ultimate delta is delta_3. The introduction of y in the proof, then the fixing of y, is nothing but an intermediary step along our ultimate goal, which was to prove that we can make f(x)/g(x) as close as we want to L by taking x sufficiently close to a; notice that there is no y in this statement. I hope this helps, and if it does not, then all I can say is keep thinking about it. :-)
@@slcmathpc I've been thinking about it more and I think I might be starting to get it. Thank you for your response.
I don't know if you still read comments but I really appreciate the way you teach concepts in such an easy manner and also the fact that your course material is free on your website. Thank you so much it really makes it easy on us High school students.
It gives me hope to know that some high school students are curious enough to seek to truly understand calculus. Thank you!
Hi, I started by watching your video about Taylor expansions, I ended up watching more than 20 videos of yours. I immediatly fell in love with how u explain, everything is clear and concise.
Big appreciation for your work and dedication.
Thank you.
But since you are fixing y, this construction doesn't force c to approach a, so you're never taking a limit of f'/g'.
You need to think of "approaching" as "can be made arbitrarily close to", which is the more mature framing of convergence. Since we can take y arbitrarily close to a, and we take x even closer to a, then this forces c to also be arbitrarily close to a. This is usually the first time people studying calculus must confront this slightly more sophisticated framing of convergence. If you ever take Advanced Calculus or Real Analysis, then you will learn the fully rigourous framing of convergence, which is the one intuitively presented in this video. Keep it up and you may one day enjoy the old epsilon-delta. ;-)
@@slcmathpc but when you send f(y)/f(x) to 0 (and similarly for g(y)/g(x)), you are fixing y and letting x approach a, and moreover, by doing so you're never forcing c to approach a. So to make this a full proof you have to justify sending f(y)/f(x) to 0 while letting BOTH y and x approach a. This can be done, but it takes a little bit more work.
See my previous comment. ;-)
In all seriousness though, you may only fully get it after learning and understanding the rigourous treatment of convergence, which I am only introducing and presenting here in an intuitive fashion. In the given proof, it means that by taking a sufficiently small open interval around a, we can make f'(c)/g'(c) arbitrarily close to L for all values of c in the given interval; this is by assumption. Then, by Cauchy's Mean Value Theorem, we can show that f(x)/g(x) can also be made arbitrarily close to L, since it can be made to be arbitrarily close to f'(c)/g'(c), for some value of c in the given interval.
@@slcmathpc I am a professional mathematician, so trust me, I understand the rigorous definition of convergence. I came across your video as I was having some trouble proving the infinity/infinity case of L'Hospital's rule. In particular, the issue I was having is precisely the issue I am bringing up with your proof (as I was proving it essentially the same way). I was eventually able to overcome it, but it took a little work. In any case, perhaps you should read my comment a bit more carefully before dismissing it. By the way, I think your videos are great, and I am not trying to be a jerk about this. I just noticed this subtlety in the proof and I thought I would bring it to your attention.
Ah well, then if you throw in an epsilon-delta in my previous comment, you can easily have yourself a fully rigourous proof of the result! By the way, this video is intended for students taking their first course in differential calculus, so arriving at a solid geometric understanding of the result is the objective. When looking at the result through this geometric lens, it appears so simple and intuitive, and it is a shame that such a beautiful demonstration is usually not presented in introductory calculus courses. Have fun completing the proof for yourself. :-)
Top five teachers on youtube!!!!!!!
Thank you!
How i understand why its not possible to apply lopital rule to
lim(x->∞)(x*sin(x)/(x*cos(x)))
Bc they give spiral
They dont go to line when x is inf
Since xsin(x) does not tend to infinity (because of the perpetual oscillation) as x does tend to infinity, then l'Hospital's rule does not apply. The same goes for xcos(x).
Hi, I was wondering how useful you find making UA-cam videos is when it comes to you learning something and if you recommend I persue the same path?
I was especially curious in your case since I'm currently in cegep and from your playlist codes I'm assuming you were part of the same course carriculum as I am taking.
That is an interesting question. In general, striving to reach a state where you can clearly and effectively explain a topic will certainly strengthen and deepen your own understanding. That being said, given that you are a young person, I would encourage you to learn a topic as quickly as you can and then to move on to a new topic. You should strive to accelerate your intellectual development. Along the way, you will certainly come accross topics that are either uniformly poorly explained or for which the explanations provided are simply not satisfactory to you. This is where I would encourage you to struggle with the ideas and try to come up with truly satisfying explanations. For those types of topics, I would encourage you to create instructional videos or notes and to share them on UA-cam. This will not only benefit you, but also many others and your content will be somewhat original. There is not much value for yourself or for others in simply repeating clear explanations of a topic that are already widely available.
On a side note, you guessed correctly as I am teaching mathematics in Cegep. I created these videos for my students and decided to make them and my course packs widely available on UA-cam. If you ever take a peek at my course packs available on my course web page, you can find out where I teach.
Good luck with your studies and finding your own path!
@@slcmathpc
Hi, thanks again for the detailed explanation I will keep that in mind. Furthermore, I'm assuming you studied math in university? If I may ask I am wondering how you found it because I'm either considering going into physics and compsci or physics and math.
Yes, I did pursue mathematics at the undergraduate and graduate level. Now, given that I do not know you, my advice will be somewhat generic. You seem to have a penchant for physics and given how much mathematics and modern physics are intertwined, you will definitely learn plenty of math by studying physics. Given that it is 2021, learning about computer science as well will open up a vast number of opportunities to you, so I would suggest you go into physics and computer science. Please, do weigh this advice appropriately since, once again, I do not know you. Best of luck!
@@slcmathpc Thanks yes those were the main reasons I myself had been told and was leaning towards since I am more of a physics person. Furthermore I was curious whether you had any good books I can get in order to advance my proof techniques, especially in linear algebra as I am teaching myself that now before the semester starts. Proofs are more foreign to me as I have not been properly taught and linear tends to be more proof heavy than cal 2.
I don't have any good book recommendation for linear algebra at the moment, but might I suggest my playlist and course pack on linear algebra (Math NYC, the Science version.) I do cover all the proofs that are accessible at this level and some of my problem sheets do contain complete solutions/proofs to some of the core theoretical related ideas/problems. A small number of proofs are omitted as they are beyond the scope of the course. This material should be useful as in the past, some of my students essentially completed the course on their own. Good luck!
Hi! I am confused by the final step of the proof that is why g(y)/g(x) is not equal to 1? As x tends to y, why g(y) not equal to infinity? Is this a rigorous proof? Thank you so much.
Have a second look; we first fix y as close as we want to a, and then let x tend to a, not y. Since we have fixed y, g(y) is constant and as we let x tend to a, g(x) tends to infinity by assumption, thus g(y)/g(x) tends to zero. :-)
@@slcmathpc Hi！After watching this video, I am confused that why we can say f'(c)/g'(c) has the same limits as f'(x)/g'(x) as c is more far away from a than x? Why they have the same limits?
Convergence means that an expression can be made arbitrarily close to the stated limit. By assumption, f'(x)/g'(x) converges to L as x approaches a, and notice that the variable x here can be replaced by any other variable different than a. As a first step, we take y to be arbitrarily close to a, so c is also arbitrarily close to a from our setup, which implies that f'(c)/g'(c) is arbitrarily close to L. As a second step, by letting x approach a, we see that we can make f(x)/g(x), with the help of Cauchy's Mean Value Theorem, arbitrarily close to f'(c)/g'(c), which is itself arbitrarily close to L. In this two-step process, we can see that f(x)/g(x) can be made arbitrarily close to L by taking x sufficiently close to a. This completes the argument that f(x)/g(x) converges to L as x approaches a. Cheers! :-)
Can be this case reduced to the 0/0 case ?
All of the so called "proofs" that I have ever seen that claim that the infinity over infinity case can be reduced to the zero over zero case are all intrinsically fallacious, meaning that they make assumptions that they cannot justify. This proof right here is the real deal. :-)
Plz make a video on integration too ☺☺
I have an entire playlist covering integral calculus. You can also download the course pack from my web page; the link is in the description of this video. :-)
Plz do video in sequences also plz
I have a playlist on differential calculus. ;-)
Cauchy mvt + g(x)=x ==> mvt
Yup.